The velocity of a particle is given by the equation v= 4t+cos4t where t is the time in seconds and v is the velocity in m s ^-1. Find the time t when the particle is no longer accelerating for the interval 0≤t≤2.

This problem tests two key sub-topics in calculus, kinematics and the chain rule. Firstly, you must realise that the derivative of a velocity function will give you the acceleration function. So by finding this derivative, you can then equal the equation to zero (As we need to find the time at which the acceleration is zero).

However, in the velocity equation, you will notice that there is a composite function that cannot be derived in a normal fashion. In your IB Maths SL formula booklet, you will find the chain rule formula. In other words, this is simply (f ° g)' = (f' ° g) ° g'. By looking at the booklet, you will also find that the derivative of cos(x) is -sin(x).

Now, time to actually tackle the maths.

In this case, g=4t ----> g'(t)=4 f(x)= cos(t) ----> f'(t)=-sin(t)

Combining the two you get

-sin(4t) ° 4 = -4sin(4t).

Then, you need to add the other part of the velocity equation and derive that, the non-composite part 4t.

Combining this gives you

v'(t)=4-4sin(4t)=a(t)

Then finally, you can equate the equation to zero and solve it. As this is a trigonometric equation, you'll need to remember your sine values.

0=4-4sin(4t) -4=-4sin(4t) -4/-4=sin(4t) 1=sin(4t)

At this stage, you know that sin is equal to one at π/2, so 4t=π/2

t=π/8 seconds.